And if this functor is representable, we say that a fine moduli space exists for this moduli problem, and even if it is not, if a certain one-one correspondence between points and objects is true over algebraically closed fields, and if a certain universal property for this is satisfied, then a coarse moduli space exists.

I suppose the above is the agreed standard terminology. Please correct me if I am wrong.

Now, the problem is that in the above definition of a moduli problem, the notion of a "functor classifying isomorphism classes of a certain type of object" is vague. We could have curves with marked points, other types of varieties with extra conditions, bundles, and so on. If we on the other hand relax the criteria and allow just any functor, then the definition becomes too broad,and any scheme will be a fine moduli space for its functor of points.

So is there a better definition, or is this all one can say? Pardon me if this was a stupid question.

2 Answers
2

Since not many people have had anything to say, I thought I might make a few remarks. But beware that this is all what I've passively picked up over the years---it's not the result of an actual study of things.

I think the right definition of a moduli problem is a fibered category $p:E\to B$. This should be thought of as a family of categories parametrized by $B$, just like you think of a map $X \to S$ of spaces as being a family of spaces parametrized by $S$. Here is an example: $E$ is the category whose objects are maps of schemes $X\to S$ making $X$ a family of elliptic curves over $S$ (i.e. an abelian scheme of relative dimension 1), and whose maps are the (hopefully) evident cartesian squares; $B$ is the category of schemes, and the functor $E\to B$ sends an object $X\to S$ to $S$. The fibered structure is given by pull back: given $X\to S$ in $E$ and a map $S'\to S$ in $B$, we get the object $X\times_S S'\to S'$ (which maps under $p$ to $S'$). If we let $E_S$ denote the fiber of this fibered category over an object $S$, then in this example, $E_S$ is the category of families of elliptic curves parametrized by $S$. Thus the fibered category encodes the data of all possible families of elliptic curves and how they behave under base change. So I hope my point that the fibered category is the moduli problem seems reasonable.

You then say the fibered category is representable if there is an object $U$ of $E$ such that for any $S$ in $B$, the pull back functor from the discrete category consisting of the set $\mathrm{Hom}(S,p(U))$ to the category $E_S$ is an equivalence. This is pretty unlikely. For instance, it implies that each category $E_S$ is discrete---all maps are isomorphisms and all automorphisms are the identity. This is certainly not the case with the elliptic curve example. Every elliptic curve has a nontrivial automorphism given by the inverse map with respect to the group structure.

Another version of representability of a fibered category is the following. Let $F:B\to\mathrm{Sets}$ denote the functor which sends an object $S$ to the set of isomorphism classes of objects of $E_S$. Then the moduli problem is (weakly?) representable if the functor $F$ is representable. This definition is surely weaker in general than the one above, but it is often equivalent in the examples that people look at in algebraic geometry. For example, the two are probably equivalent if each $E_S$ is a discrete category.

If the moduli problem is not representable, then you get into other issues, such as whether you have effective descent with respect to some topology on $B$ (i.e. $E$ is a "sheaf of categories" over $B$), and if so, stack-theoretic issues, such as whether $E$ can be represented by a category object in $E$, and if so, whether it's a groupoid object.

One more thing: you definitely want every scheme to be the fine moduli space for its functor of points, or rather for the fibered category of families of maps to it. That's a mandatory requirement, not a defect!
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JBorgerFeb 3 '10 at 11:05

That is a frown and a smile, for the record.
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Harry GindiFeb 3 '10 at 11:44

This was helpful. Is it possible to impose a sheaf condition on a fibered category so that descent is possible?
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AnweshiFeb 3 '10 at 19:47

That's an interesting question. I don't know the answer. I think I might have heard the word "stackification", so maybe someone has worked it out. But it's probably not completely easy/ Even the usual sheafification functor is not so simple -- in fact it's pretty much the only non-simple thing in general sheaf theory
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JBorgerFeb 3 '10 at 20:53

The closest I can come to a "definition" of a moduli functor is "a quotient of a subfunctor of the functor of points of a Quot scheme". This is a purely sociological definition; one can study other objects, but it might be hard and lonely.